Probability (Devore Chapter Two)

Transcription

1 Probability (Devore Chapter Two) Probability Winter Contents 1 Axiomatic Probability Outcomes and Events Rules of Probability Venn Diagrams Assigning Probabilities Counting Techniques Ordered Sequences Permutations and Combinations Conditional Probabilities and Tree Diagrams Example: Odds of Winning at Craps Definition of Conditional Probability Bayes s Theorem Approach Considering a Hypothetical Population Approach Using Axiomatic Probability Copyright 2011, John T. Whelan, and all that 1

2 Tuesday 6 December Axiomatic Probability Many of the rules of probability appear to be self-evident, but it s useful to have a precise language in which they can be described. To that end, Devore develops with some care a mathematical theory of probability. We ll mostly summarize the key definitions and results here. 1.1 Outcomes and Events Devore defines probability in terms of an experiment which can have one of a set of possible outcomes. The sample space of an experiment, written S, is the set of all possible outcomes. An event is a subset of S, a set of possible outcomes to the experiment. Special cases are: The null event is an event consisting of no outcomes (the empty set) A simple event consists of exactly one outcome A compound event consists of more than one outcome The sample space S itself an event, of course. One example of an experiment is flipping a coin three times. The outcomes in that case are HHH, HHT, HT H, HT T, T HH, T HT, T T H, and T T T. Possible outcomes include: Exactly two heads: {HHT, HT H, T HH} The first flip is heads: {HHH, HHT, HT H, HT T } The second and third flips are the same: {HHH, HT T, T HH, T T T } Another example is a game of craps, in which: if a 2, 3 or 12 is rolled on the first roll, the shooter loses if a 7 or 11 is rolled on the first roll, the shooter wins if a 4, 5, 6, 8, 9, or 10 is rolled on the first roll, the dice are rolled again until the either that number or a 7 comes up, in which case the shooter wins or loses, respectively. In this case there are an infinite number of outcomes in S, some of which are: 2, 3, 7, 11, 12, 4 4, 4 7, 5 5, 5 7, 6 6, 6 7, 8 8, 8 7, 9 9, 9 7, 10 10, 10 7, 4 2 4, 4 3 4, 4 5 4, 4 6 4,.... Possible events include: the shooter wins {7, 11, 4 4, 5 5, 6 6, 8 8,...}; the shooter loses {2, 3, 12, 4 7,...}; the dice are thrown exactly once {2, 3, 7, 11, 12}, etc. Since an event is a set of outcomes, we can use all of the machinery of set theory, specifically: The complement A of an event A, is the set of all outcomes in S which are not in A. 2

3 The union A B of two events A and B, is the set of all outcomes which are in A or B, including those which are in both. The intersection A B is the set of all outcomes which are in both A and B. In the case of coin flips, if the events are A = {HHT, HT H, T HH} (exactly two heads) and B = {HHH, HHT, HT H, HT T } (first flip heads), we can construct, among other things, A = {HHH, HT T, T HT, T T H, T T T } A B = {HHH, HHT, HT H, HT T, T HH} A B = {HHT, HT H} Another useful definition is that A and B are disjoint or mutually exclusive events if A B =. Note that the trickiest part of many problems is actually keeping straight what the events are to which you re assigning probabilities! 1.2 Rules of Probability Having formally defined what we mean by an event, we can proceed to define the probability of that event, which we think of as the chance that it will occur. Devore starts with three axioms 1. For any event A, P (A) 0 2. P (S) = 1 3. Given an infinite collection A 1, A 2, A 3,... of disjoint events, ( ) P (A 1 A 2 A 3 ) = P A i = P (A i ) (1.1) From there he manages to derive a bunch of other sensible results, such as 1. For any event A, P (A) 1 2. P ( ) = 0 3. P (A ) = 1 P (A) One useful result concerns the probability of the union of any two events. Since A B = (A B ) (A B) (A B), the union of three disjoint events, i=1 P (A B) = P (A B ) + P (A B) + P (A B) (1.2) On the other hand, A = (A B ) (A B) and B = (A B) (A B), so i=1 P (A) = P (A B ) + P (A B) P (B) = P (A B) + P (A B) (1.3a) (1.3b) 3

4 which means that P (A) + P (B) = P (A B ) + 2P (A B) + P (A B) = P (A B) + P (A B) (1.4) so P (A B) = P (A) + P (B) P (A B) (1.5) 1.3 Venn Diagrams University Website Note that this is a Venn diagram illustrating the relationship between two sets of things rather than two sets of outcomes 4

5 We can often gain insight into addition of probabilities with Venn diagrams, in which events are represented pictorially as regions in a plane. For example, here the two overlapping circles represent the events A and B: The intersection of those two events is shaded here: 5

6 The union the two events is shaded here: Thus we see that if we add P (A) and P (B) by counting all of the outcomes in each circle, we ve double-counted the outcomes in the overlap, which is why we have to subtract P (A B) in (1.5) 1.4 Assigning Probabilities The axioms of probability let us relate the probabilities of different events, but they don t tell us what those probabilities should be in the first place. If we have a way of assigning probabilities to each outcome, and therefore each simple event, then we can use the sum rule for disjoint events to write the probability of any event as the sum of the probabilities of the simple events which make it up. I.e., P (A) = P (E i ) (1.6) E i in A One possibility is that each outcome, i.e., each simple event, might be equally likely. In that case, if there are N outcomes total, the probability of each of the simple events is P (E i ) = 1/N (so that N i=1 P (E i) = P (S) = 1), and in that case P (A) = E i in A 1 N = N(A) N (1.7) where N(A) is the number of outcomes which make up the event A. Note, however, that one has to consider whether it s appropriate to take all of the outcomes to be equally likely. For instance, in our craps example, we considered each roll, e.g., 6

7 2 and 4 to be its own outcome. But you can also consider the rolls of the individual dice, and then the two dice totalling 4 would be a composite event consisting of the outcomes (1, 3), (2, 2), and (3, 1). For a pair of fair dice, the 36 possible outcomes defined by the numbers on the two dice taken in order (suppose one die is green and the other red) are equally likely outcomes. 2 Counting Techniques 2.1 Ordered Sequences We can come up with 36 as the number of possible results on a pair of fair dice in a couple of ways. We could make a table which is also useful for counting the number of occurrences of each total. Or we could use something called a tree diagram: 7

8 This works well for counting a small number of possible outcomes, but already with 36 outcomes it is becoming unwieldy. So instead of literally counting the possible outcomes, we should calculate how many there will be. In this case, where the outcome is an ordered pair of numbers from 1 to 6, there are 6 possibilities for the first number, and corresponding to each of those there are 6 possibilities for the second number. So the total is 6 6 = 36. More generally, if we have an ordered set of k objects, with n 1 possibilities for the first, n 2 for the second, etc, the number of possible ordered k-tuples is n 1 n 2... n k, which we can also write as k n k. (2.1) i=1 2.2 Permutations and Combinations Consider the probability of getting a poker hand (5 cards out of the 52-card deck) which consists entirely of hearts. 1 Since there are four different suits, you might think the odds are (1/4)(1/4)(1/4)(1/4)(1/4) = (1/4) 5 = 1/4 5. However, once a heart has been drawn on the first card, there are only 12 hearts left in the deck out of 51; after two hearts there are 11 out of 50, etc., so the actual odds are ( ) ( ) ( ) ( ) ( ) P ( ) = (2.2) This turns out not to be the most effective way to calculate the odds of poker hands, though. (For instance, it s basically impossible to do a card-by-card accounting of the probability of getting a full house.) Instead we d like to take the approach of counting the total number of possible five-card hands (outcomes) and then counting up how many fall into a particular category (event). The terms for the quantities we will be interested in are permutation and combination. First, let s consider the number of possible sequences of five cards drawn out of a deck of 52. This is the permutation number of permutations of 5 objects out of 52, called P 5,52. The first card can be any of the 52; the second can be any of the remaining 51; the third can be any of the remaining 50, etc. The number of permutations is In general P 5,52 = (2.3) k 1 P k,n = n(n 1)(n 2) (n k + 1) = (n l). (2.4) Now, there is a handy way to write this in terms of the factorial function. Remember that the factorial is defined as n n! = n(n 1)(n 1) (2)(1) = l (2.5) 1 This is, hopefully self-apparently, one-quarter of the probability of getting a flush of any kind. l=0 l=1 8

9 with the special case that 0! = 1. Then we can see that n! (n k)! = n(n 1)(n 2) (n k + 1) (n k) (n k 1) (2) (1) (n k) (n k 1) (2) (1) = P k,n (2.6) Note in particular that the number of ways of arranging n items is P n,n = n! (n n)! = n! 0! = n! (2.7) Now, when we think about the number of different poker hands, actually we don t consider the cards in a hand to be ordered. So in fact all we care about is the number of ways of choosing 5 objects out of a set of 52, without regard to order. This is the number of combinations, which is sometimes written C 5,52, but which we ll write as ( ) 52 5, pronounced 52 choose 5. When we counted the number of different permutations of 5 cards out of 52, we actually counted each possible hand a bunch of times, once for each of the ways of arranging the cards. There are P 5,5 = 5! different ways of arranging the five cards of a poker hand, so the number of permutations of 5 cards out of 52 is the number of combinations times the number of permutations of the 5 cards among themselves: ( ) 52 P 5,52 = P 5,5 (2.8) 5 The factor of P 5,5 = 5! is the factor by which we overcounted, so we divide by it to get ( ) 52 = P 5,52 = 52! = (2.9) 5 P 5,5 47!5! or in general ( ) n = k n! (n k)!k! (2.10) So to return to the question of the odds of getting five hearts, there are ( ) 52 5 different poker hands, and ( ) 13 5 different hands of all hearts (since there are 13 hearts in the deck), which means the probability of the event A = is P (A) = N(A) ( 13 ) N = 5 ( 52 ) = = 13!47! = (13)(12)(11)(10)(9) (2.11) 8!52! (52)(51)(50)(49)(48) 5 13! 8!5! 52! 47!5! which is of course what we calculated before. Numerically, P (A) , while 1/ The odds of getting any flush are four times the odds of getting an all heart flush, i.e., Actually, if we want to calculate the odds of getting a flush, we have over-counted somewhat, since we have also included straight flushes, e.g., If we want to 9

10 count only hands which are flushes, we need to subtract those. Since aces can count as either high or low, there are ten different all-heart straight flushes, which means the number of different all-heart flushes which are not straight flushes is ( ) = 13! 10 = = 1277 (2.12) 5 8!5! and the probability of getting an all-heart flush is , or for any flush. Exercise: work out the number of possible straights and therefore the odds of getting a straight. Practice Problems 2.5, 2.9, 2.13, 2.17, 2.29, 2.33,

11 Thursday 8 December Conditional Probabilities and Tree Diagrams Conditional Risk Example: Odds of Winning at Craps Although there are an infinite number of possible outcomes to a craps game, we can still calculate the probability of winning. First, the sample space can be divided up into mutually exclusive events based on the result of the first roll: 11

12 Event Probability Result of game , 3 or 12 on 1st roll = % lose or 11 on 1st roll = % win or 10 on 1st roll = %??? or 9 on 1st roll = %??? or 8 on 1st roll = %??? The last three events each contain some outcomes that correspond to winning, and some that correspond to losing. We can figure out the probability of winning if, for example, you roll a 4 initially. Then you will win if another 4 comes up before a 7, and lose if a 7 comes up before a 4. On any given roll, a 7 is twice as likely to come up as a 4 (6/36 vs 3/36), so the odds are 6/9 = 2/3 66.7% that you will roll a 7 before a 4 and lose. Thus the odds of losing after starting with a 4 are 66.7%, while the odds of winning after starting with a 4 are 33.3%. The same calculation applies if you get a 10 on the first roll. This means that the 6/ % probability of rolling a 4 or 10 initially can be divided up into a 4/ % probability to start with a 4 or 10 and eventually lose, and a 2/36 5.6% probability to start with a 4 or 10 and eventually win. We can summarize this branching of probabilities with a tree diagram: The probability of winning given that you ve rolled a 4 or 10 initially is an example of a conditional probability. If A is the event roll a 4 or 10 initially and B is the event win 12

13 the game, we write the conditional probability for event B given that A occurs as P (B A). We have argued that the probability for both A and B to occur, P (A B), should be the probability of A times the conditional probability of B given A, i.e., P (A B) = P (B A)P (A) (3.1) We can use this to fill out a table of probabilities for different sets of outcomes of a craps game, analogous to the tree diagram. A P (A) B P (B A) P (A B) = P (B A)P (A) 2, 3 or 12 on 1st roll.111 lose or 11 on 1st roll.222 win or 10 on 1st roll.167 lose win or 9 on 1st roll.222 lose win or 8 on 1st roll.278 lose win Since the rows all describe disjoint events whose union is the sample space S, we can add the probabilities of winning and find that P (win) (3.2) and P (lose) (3.3) 3.2 Definition of Conditional Probability We ve motivated the concept of conditional probability and applied it via (3.1). In fact, from a formal point of view, conditional probability is defined as P (B A) = P (A B) P (A). (3.4) We actually used that definition in another context above without realizing it, when we were calculating the probability of rolling a 7 before rolling a 4. We know that P (7) = 6/36 and P (4) = 3/36 on any given roll. The probability of rolling a 7 given that the game ends on that throw is P (7 7 4) = P (7) P (7 4) = P (7) P (7) + P (4) = 6/36 9/36 = 6 (3.5) 9 We calculated that using the definition of conditional probability. 13

14 4 Bayes s Theorem Some of the arguments in this section are adapted from which gives a nice explanation of Bayes s theorem. The laws of probability are pretty good at predicting how likely something is to happen given certain underlying circumstances. But often what you really want to know is the opposite: given that some thing happened, what were the circumstances? The classic example of this is a test for a disease. Suppose that one one-thousandth of the population has a disease. There is a test that can detect the disease, but it has a 2% false positive rate (on average one out of fifty healthy people will test positive) and as 1% false negative rate (on average one out of one hundred sick people will test negative). The question we ultimately want to answer is: if someone gets a positive test result, what is the probability that they actually have the disease. Note, it is not 98%! 4.1 Approach Considering a Hypothetical Population The standard treatment of Bayes s Theorem and the Law of Total Probability can be sort of abstract, so it s useful to keep track of what s going on by considering a hypothetical population which tracks the various probabilities. So, assume the probabilities arise from a population of 100,000 individuals. Of those, one one-one-thousandth, or 100, have the disease. The other 99,900 do not. The 2% false positive rate means that of the 99,900 healthy individuals, 2% of them, or 1,998, will test positive. The other 97,902 will test negative. The 1% false negative rate means that of the 100 sick individuals, one will test negative and the other 99 will test positive. So let s collect this into a table: Positive Negative Total Sick Healthy 1,998 97,902 99,900 Total 2,097 97, ,000 (As a reminder, if we choose a sample of 100,000 individuals out of a larger population, we won t expect to get exactly this number of results, but the 100,000-member population is a useful conceptual construct.) Translating from numbers in this hypothetical population, we can confirm that it captures the input information: P (sick) = 100 = , 000 (4.1a) 1, 998 P (positive healthy) = =.02 99, 900 (4.1b) P (negative sick) = 1 = (4.1c) 14

15 But now we can also calculate what we want, the conditional probability of being sick given a positive result. That is the fraction of the total number of individuals with positive test results that are in the sick and positive category: P (sick positive) = 99 2, (4.2) or about 4.7%. Note that we can forego the artificial construct of a 100,000-member hypothetical population. If we divide all the numbers in the table by 100,000, they become probabilities for the corresponding events. For example, P (sick positive) = , 000 = That is the approach of the slightly more axiomatic (and general) method described in the next section. 4.2 Approach Using Axiomatic Probability The quantity we re looking for (the probability of being sick, given a positive test result) is a conditional probability. To evaluate it, we need to define some events about which we ll discuss the probability. First, consider the events A 1 individual has the disease A 2 individual does not have the disease (4.3a) (4.3b) These make a mutually exclusive, exhaustive set of events, i.e., A 1 A 2 = and A 1 A 2 = S. (We call them A 1 and A 2 because in a more general case there might be more than two events in the mutually exclusive, exhaustive set.) We are told in the statement of the problem that one person in 1000 has the disease, which means that P (A 1 ) =.001 P (A 2 ) = 1 P (A 1 ) =.999 (4.4a) (4.4b) (4.4c) Now consider the events associated with the test: B individual tests positive B individual tests negative (4.5a) (4.5b) (We call them B and B because we will focus more on B.) The 2% false positive and 1% false negative rates tell us P (B A 2 ) =.02 P (B A 1 ) =.01 (4.6a) (4.6b) 15

16 Note that if we want to talk about probabilities involving B, we should use the fact that to state that P (B A 1 ) + P (B A 1 ) = 1 (4.7) P (B A 1 ) = 1 P (B A 1 ) =.99 (4.8) Now we can write down the quantity we actually want, using the definition of conditional probability: P (A 1 B) = P (A 1 B) (4.9) P (B) Now, we don t actually have an expression for P (A 1 B) or P (B) yet. The fundamental things we know are P (A 1 ) =.001 P (A 2 ) =.999 P (B A 2 ) =.02 P (B A 1 ) =.99 (4.10a) (4.10b) (4.10c) (4.10d) However, we know how to calculate P (A 1 B) and P (B) from the things we do know. First, to get P (A 1 B), we can notice that we know P (B A 1 ) and P (A 1 ), so we can solve P (B A 1 ) = P (A 1 B) P (A 1 ) (4.11) for P (A 1 B) = P (B A 1 )P (A 1 ) = (.99)(.001) = (4.12) Logically, the probability of having the disease and testing positive for it is the probability of having the disease in the first place times the probability of testing positive, given that you have the disease. Since we know P (B A 2 ) and P (A 2 ) we can similarly calculate the probability of not having the disease but testing positive anyway: P (A 2 B) = P (B A 2 )P (A 2 ) = (.02)(.999) = (4.13) But now we have enough information to calculate P (B), since the overall probability of testing positive for the disease has to be the probability of having the disease and testing positive plus the probability of not having the disease and testing positive: P (B) = P (A 1 B) + P (A 2 B) = = (4.14) This is an application of the Law of total probability, which says, in the more general case of k mutually exclusive, exhaustive events A 1,..., A k, P (B) = P (B A 1 ) + + P (B A k ) = P (B A 1 )P (A 1 ) + + P (B A k )P (A k ) (4.15) 16

17 Now we are ready to calculate P (A 1 B) = P (A 1 B) P (B) = (4.16) So only about 4.7% of people who test positive have the disease. It s a lot more than one in a thousand, but a lot less than 99%. This is an application of Bayes s theorem, which says P (A 1 B) = P (A 1 B) P (B) = P (B A 1 )P (A 1 ) P (B A 1 )P (A 1 ) + P (B A 2 )P (A 2 ) (4.17) or, for k mutually exclusive, exhaustive alternatives, P (A 1 B) = P (B A 1)P (A 1 ) k i=1 P (B A i)p (A i ) (4.18) Practice Problems 2.45, 2.59, 2.63, 2.67, 2.71, parts a & b 17